Answer
$\mu_k = 0.54$
Work Step by Step
We can find the acceleration:
$d = \frac{1}{2}at^2$
$a = \frac{2d}{t^2}$
$a = \frac{(2)(2.5~m)}{(4.0~s)^2}$
$a = 0.3125~m/s^2$
We can find the coefficient of kinetic friction:
$mg~sin~\theta-mg~\mu_k~cos~\theta = ma$
$g~sin~\theta-g~\mu_k~cos~\theta = a$
$g~\mu_k~cos~\theta = g~sin~\theta - a$
$\mu_k = \frac{g~sin~\theta - a}{g~cos~\theta}$
$\mu_k = \frac{(9.8~m/s^2)~sin~30^{\circ} - (0.3125~m/s^2)}{(9.8~m/s^2)~cos~30^{\circ}}$
$\mu_k = 0.54$
